Question

The sum of digits of a 2-digit number is 11. If the number obtained by reversing the
digits is 9 less than the original number, find the number.
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Answer 1

Let the digit at ones place be y and at tens place be x.

So,

Number = 10x + y

According to the question:

x + y = 11 ...(1)

10y + x = 10x + y + 9

10y - y + x - 10x = 9

9y - 9x = 9

x - y = 1 ...(2)

Subtract (1) from (2)

x - y - x - y = 1 - 11

-2y = -10

2y = 10

y = 5

Putting value of y in (1)

x + y = 11

x + 5 = 11

x = 11 - 5

x = 6

Original number = 10x + y

=> 10(6) + 5

=> 60 + 5

=> 65

Hence, The number is 65